When finding global extrema of single-variable functions on a closed interval, we check critical values over that interval and then evaluate the function at the endpoints. Let’s briefly recall how this works.
Extreme Value Theorem (Single-Variable)
A continuous function [latex]f(x)[/latex] on a closed interval [latex][a,b][/latex] attains an absolute maximum value at some point and an absolute minimum value at some point. These values are the largest and smallest values found among:
The values of [latex]f[/latex] at the critical values within [latex][a,b][/latex]
The values of [latex]f[/latex] at the endpoints of [latex][a,b][/latex]
For functions of two variables, the closed interval is replaced by a closed, bounded set. A set is bounded if all points in that set can be contained within a ball (or disk) of finite radius. We need to find the critical points inside the set and calculate the corresponding critical values. Then, we find the maximum and minimum values on the boundary of the set. The largest function value corresponds to the global maximum and the smallest to the absolute minimum.
First, we need to be assured that such values exist.
extreme value theorem
A continuous function [latex]f(x, y)[/latex] on a closed and bounded set [latex]D[/latex] in the plane attains an absolute maximum value at some point of [latex]D[/latex] and an absolute minimum value at some point of [latex]D[/latex].
Now that we know continuous functions on closed, bounded sets attain their extreme values, we need to know how to find them.
finding absolute extrema
Assume [latex]z=f(x, y)[/latex] is a differentiable function of two variables defined on a closed, bounded set [latex]D[/latex]. Then [latex]f[/latex] will attain the absolute maximum value and the absolute minimum value, which are the largest and smallest values found among:
The values of [latex]f[/latex] at the critical points of [latex]f[/latex] in [latex]D[/latex]
The values of [latex]f[/latex] on the boundary of [latex]D[/latex]
The proof follows directly from the extreme value theorem and Fermat’s theorem. If either extremum is not on the boundary of [latex]D[/latex], then it’s at an interior point of [latex]D[/latex]. But an interior point [latex](x_0, y_0)[/latex] that’s an absolute extremum is also a local extremum, so [latex](x_0, y_0)[/latex] is a critical point by Fermat’s theorem. Therefore, the only possible values for global extrema are the extreme values on the interior or boundary of [latex]D[/latex].
Finding Absolute Extrema on a Closed, Bounded Set
Let [latex]z=f(x, y)[/latex] be a continuous function defined on a closed, bounded set [latex]D[/latex], and assume [latex]f[/latex] is differentiable on [latex]D[/latex]. To find the absolute maximum and minimum values:
Find the critical points of [latex]f[/latex] in [latex]D[/latex]
Calculate [latex]f[/latex] at each critical point
Determine the maximum and minimum values of [latex]f[/latex] on the boundary of [latex]D[/latex]
Compare all values from steps 2 and 3—the maximum and minimum will occur at one of these values
Finding extrema on the boundary can be challenging. If the boundary is a rectangle or set of straight lines, parameterize the line segments and determine the maxima on each segment. The same approach works for circles and ellipses.
If the boundary is a more complicated curve defined by [latex]g(x, y)=c[/latex], and the first-order partial derivatives of [latex]g[/latex] exist, the method of Lagrange multipliers can help determine extrema on the boundary.
Use the problem-solving strategy for finding absolute extrema of a function to determine the absolute extrema of each of the following functions:
[latex]f(x, y)=x^{2}-2xy+4y^{2}-4x-2y+24[/latex] on the domain defined by [latex]0\leq x\leq 4[/latex] and [latex]0\leq y\leq 2[/latex].
[latex]g(x, y)=x^{2}+y^{2}+4x-6y[/latex] on the domain defined by [latex]x^{2}+y^{2}\leq 16[/latex].
a. Using the problem-solving strategy, step 1 involves finding the critical points of [latex]f[/latex] on its domain. Therefore, we first calculate [latex]f_x(x, y)[/latex] and [latex]f_y(x, y)[/latex], then set them each equal to zero:
The solution to this system is [latex]x=3[/latex] and [latex]y=1[/latex]. Therefore [latex](3, 1)[/latex] is a critical point of [latex]f[/latex]. Calculating [latex]f(3, 1)[/latex] gives [latex]f(3, 1)=17[/latex]. The next step involves finding the extrema of [latex]f[/latex] on the boundary of its domain. The boundary of its domain consists of four line segments as shown in the following graph:
Figure 1. Graph of the domain of the function [latex]\small{f(x,y)=x^{2}-2xy+4y^{2}-4x-2y+24}[/latex].
[latex]L_1[/latex] is the line segment connecting [latex](0, 0)[/latex] and [latex](4, 0)[/latex], and it can be parameterized by the equations [latex]x(t)=t[/latex], [latex]y(t)=0[/latex] for [latex]0\leq t\leq 4[/latex]. Define [latex]g(t)=f(x(t), y(t))[/latex]. This gives [latex]g(t)=t^{2}-4t+24[/latex]. Differentiating [latex]g[/latex] leads to [latex]g'(t)=2t-4[/latex]. Therefore, [latex]g[/latex] has a critical value at [latex]t=2[/latex], which corresponds to the point [latex](2, 0)[/latex]. Calculating [latex]f(2, 0)[/latex] gives the [latex]z[/latex]-value [latex]20[/latex].
[latex]L_2[/latex] is the line segment connecting [latex](4, 0)[/latex] and [latex](4, 2)[/latex], and it can be parameterized by the equations [latex]x(t)=4[/latex], [latex]y(t)=t[/latex] for [latex]0\leq t\leq 2[/latex]. Again, define [latex]g(t)=f(x(t), y(t))[/latex]. This gives [latex]g(t)=4t^{2}-10t+24[/latex]. Then, [latex]g'(t)=8t-10[/latex]. [latex]g[/latex] has a critical value at [latex]t=\frac{5}{4}[/latex], which corresponds to the point [latex](4, \frac{5}{4})[/latex]. Calculating [latex]f(4, \frac{5}{4})[/latex] gives the [latex]z[/latex]-value [latex]17.75[/latex].
[latex]L_3[/latex] is the line segment connecting [latex](0, 2)[/latex] and [latex](4, 2)[/latex], and it can be parameterized by the equations [latex]x(t)=t[/latex], [latex]y(t)=2[/latex] for [latex]0\leq t\leq 4[/latex]. Again, define [latex]g(t)=f(x(t), y(t))[/latex]. This gives [latex]g(t)=t^{2}-8t+36[/latex]. The critical value corresponds to the point [latex](4, 2)[/latex]. So, calculating [latex]f(4, 2)[/latex] gives the [latex]z[/latex]-value [latex]20[/latex].
[latex]L_4[/latex] is the line segment connecting [latex](0, 0)[/latex] and [latex](0, 2)[/latex], and it can be parameterized by the equations [latex]x(t)=0[/latex], [latex]y(t)=t[/latex] for [latex]0\leq t\leq 2[/latex]. This time [latex]g(t)=4t^{2}-2t+24[/latex] and the critical value [latex]t=\frac{1}{4}[/latex]corresponds to the point [latex](0, \frac{1}{4})[/latex]. Calculating [latex]f(0, \frac{1}{4})[/latex] gives the [latex]z[/latex]-value [latex]23.75[/latex]. We also need to find the values of [latex]f(x, y)[/latex] at the corners of its domain. These corners are located at [latex](0, 0)[/latex], [latex](4, 0)[/latex], [latex](4, 2)[/latex], and [latex](0, 2)[/latex]:
[latex]\hspace{5cm}\begin{align} f(0,0)&=(0)^2-2(0)(0)+4(0)^2-4(0)-2(0)+24=24 \\ f(4,0)&=(4)^2-2(4)(0)+4(0)^2-4(4)-2(0)+24=24 \\ f(4,2)&=(4)^2-2(4)(2)+4(2)^2-4(4)-2(2)+24=20 \\ f(0,2)&=(0)^2-2(0)(2)+4(2)^2-4(0)-2(2)+24=36. \end{align}[/latex]
The absolute maximum value is [latex]36[/latex], which occurs at [latex](0, 2)[/latex], and the global minimum value is [latex]17[/latex], which occurs at [latex](3, 1)[/latex] as shown in the following figure.
Figure 2. The function [latex]\small{f(x,y)}[/latex] has one global minimum and one global maximum over its domain.
b. Using the problem-solving strategy, step 1 involves finding the critical points of [latex]g[/latex] on its domain. Therefore, we first calculate [latex]g_x(x, y)[/latex] and [latex]g_y(x, y)[/latex], then set them each equal to zero:
Setting them equal to zero yields the system of equations
[latex]\hspace{8cm}\begin{align} 2x+4&=0 \\ 2y-6&=0. \end{align}[/latex]
The solution to this system is [latex]x=-2[/latex] and [latex]y=3[/latex]. Therefore, [latex](-2, 3)[/latex] is a critical point of [latex]g[/latex]. Calculating [latex]g(-2, 3)[/latex], we get
The next step involves finding the extrema of [latex]g[/latex] on the boundary of its domain. The boundary of its domain consists of a circle of radius [latex]4[/latex] centered at the origin as shown in the following graph.
Figure 3. Graph of the domain of the function [latex]\small{g(x,y)=x^{2}+y^{2}+4x-6y}[/latex].
The boundary of the domain of [latex]g[/latex] can be parameterized using the functions [latex]x(t)=4\cos t[/latex], [latex]y(t)=4\sin t[/latex] for [latex]0\leq{t}\leq2\pi[/latex]. Define [latex]h(t)=g(x(t), y(t))[/latex]:
[latex]\hspace{5cm}\begin{align} h(t)&=g(x(t),y(t)) \\ &=(4\cos{t})^2+(4\sin t)^2+4(4\cos{t})-6(4\sin t) \\ &=16\cos^2{t}+16\sin^2{t}+16\cos{t}-24\sin t \\ &=16+16\cos{t}-24\sin t.\end{align}[/latex]
Setting [latex]h'(t)=0[/latex] leads to
[latex]\hspace{5cm}\begin{align}-16\sin{t}-24\cos{t}&=0 \\-16\sin{t}&=24\cos{t} \\\frac{-16\sin{t}}{-16\cos{t}}&=\frac{24\cos{t}}{-16\cos{t}} \\\tan{t}&=-\frac32\end{align}[/latex]
This equation has two solutions over the interval [latex]0\leq{t}\leq2\pi[/latex]. One is [latex]t=\pi-\arctan{(\frac32)}[/latex] and the other is [latex]t=2\pi-\arctan{(\frac32)}[/latex]. For the first angle,
[latex]\hspace{4cm}\begin{align}\sin{t}&=\sin{(\pi-\arctan{(\frac32)})}=\sin{(\arctan{(\frac32)})}=\frac{3\sqrt{13}}{13} \\\cos{t}&=\cos{(\pi-\arctan{(\frac32)})}=-\cos{(\arctan{(\frac32)})}=-\frac{2\sqrt{13}}{13}\end{align}[/latex]
Therefore, [latex]x(t)=4\cos{t}=\frac{8\sqrt{13}}{13}[/latex] and [latex]y(t)=4\sin{t}=-\frac{12\sqrt{13}}{13}[/latex], so [latex]\left(\frac{8\sqrt{13}}{13},-\frac{12\sqrt{13}}{13}\right)[/latex] is a critical point on the boundary and[latex]\hspace{4cm}\begin{align}g\left(-\frac{8\sqrt{13}}{13},\frac{12\sqrt{13}}{13}\right)&=\left(-\frac{8\sqrt{13}}{13}\right)^2+\left(\frac{12\sqrt{13}}{13}\right)^2+4\left(-\frac{8\sqrt{13}}{13}\right)-6\left(\frac{12\sqrt{13}}{13}\right) \\&=\frac{144}{13}+\frac{64}{13}-\frac{32\sqrt{13}}{13}-\frac{72\sqrt{13}}{13} \\&=\frac{208-104\sqrt{13}}{13}\approx-12.844.\end{align}[/latex]
For the second angle,
[latex]\hspace{5cm}\begin{align}\sin{t}&=\sin{(2\pi-\arctan{(\frac32)})}=-\sin{(\arctan{(\frac32)})}=-\frac{3\sqrt{13}}{13} \\\cos{t}&=\cos{(2\pi-\arctan{(\frac32)})}=\cos{(\arctan{(\frac32)})}=\frac{2\sqrt{13}}{13}.\end{align}[/latex]
Therefore, [latex]x(t)=4\cos{t}=\frac{8\sqrt{13}}{13}[/latex] and [latex]y(t)=4\sin{t}=-\frac{12\sqrt{13}}{13}[/latex], so [latex]\left(\frac{8\sqrt{13}}{13},-\frac{12\sqrt{13}}{13}\right)[/latex] is a critical point on the boundary and[latex]\hspace{4cm}\begin{align}g\left(\frac{8\sqrt{13}}{13},-\frac{12\sqrt{13}}{13}\right)&=\left(\frac{8\sqrt{13}}{13}\right)^2+\left(-\frac{12\sqrt{13}}{13}\right)^2+4\left(\frac{8\sqrt{13}}{13}\right)-6\left(-\frac{12\sqrt{13}}{13}\right) \\&=\frac{144}{13}+\frac{64}{13}+\frac{32\sqrt{13}}{13}+\frac{72\sqrt{13}}{13} \\&=\frac{208+104\sqrt{13}}{13}\approx44.844.\end{align}[/latex]The absolute minimum of [latex]g[/latex] is [latex]-13[/latex] which is attained at the point [latex](-2, 3)[/latex], which is an interior point of [latex]D[/latex]. The absolute maximum of [latex]g[/latex] is approximately equal to [latex]44.844[/latex], which is attained at the boundary point [latex]\left(\frac{8\sqrt{13}}{13},-\frac{12\sqrt{13}}{13}\right)[/latex]. These are the absolute extrema of [latex]g[/latex] on [latex]D[/latex] as shown in the following figure.
Figure 4. The function [latex]\small{f(x,y)}[/latex] has a local minimum and a local maximum.
